CHEMISTRY 2430 - Principles of Organic Chemistry I - Exam IV
April 9, 1999 - Dr. Howard Black

Name KEY

Throughout this exam, give answers only in the provided spaces (boxes or lines). Answers not in these spaces will not be graded, even if correct. Structures are always on sale for one (1) point each.

1. (12 pts) Give acceptable names for the following compounds, being sure to address alkene geometry and R/S configuration where appropriate.


 

 
2. (14 pts) For each of the following four molecules, list the properties from the list on the right which best describe it. Each item on the list can be used as often as necessary, or not at all. DON'T GUESS!!
 
a. contains no sp2 carbons
b. reacts with Ag(NH3)2+ (Tollens' reagent)
c. forms a salt with HCl
d. very easily oxidized
e. does not react with LiAlH4
f. incompatible with Grignard reagents
g. forms an alkene when treated with H+/D
h. reacts with NaBH4 to form a primary alcohol
i. cannot be oxidized without C-C bond fission
j. most basic of the four
k. can be formed from oxidation of a primary alcohol
 

 
3. (30 pts) Complete the following reactions (provide either the starting molecule, the reagent(s), or product(s). If more than one product is produced, draw them all. If one product predominates, circle it.


4. (16 pts) Indicate a synthesis of any TWO (2) of the following four molecules from benzene and bromoethane. You may use any inorganic reagents you like, but all carbon atoms must come from the two starting materials provided. Please fill in the letter of your choices in the spaces provided. Work out the details of your syntheses on the paper provided. The spaces below should contain only your finished, eraser- and crossout-free synthetic schemes.

There are many correct synthetic approaches to each target; these below are just possibilities.  For instance, a totally different sequence to afford A would be to do a Friedel-Crafts acylation reaction on benzene with CH3COCl (made easily from EtBr), and then treat the resulting PhCOCH3 with H2/cat. - reducing both the C=O and the aromatic ring.

5. (16 pts total) a. (10 pts) Phenylacetone (1-phenyl-2-propanone) features two types of acidic protons. Draw the structure in the box, including all non-aromatic protons, and label which proton group is the more acidic. Then, explain your reasoning by drawing resonance structures of the conjugate bases for each proton type. You must use resonance structures in your answer - those without resonance structures will receive a zero without further grading.


NOTE:  Take a look at the similar question on Exam IV from 1998 (also question #5); the concept is identical.  In both, loss of Hb (a methyl ketone) gives an anion stabilized by only one carbonyl, while loss of Ha gives an anion stabilized not only by a carbonyl, but by something additional.  In 1998, it was an additional C=O; in 1999 it's an aromatic ring.
However (to coin a phrase), it's the same thing!!!

b. (6 pts) The protons on the carbon directly bound to the aromatic ring become much more acidic if a nitro group is added to the ring (let's say para). Explain how it increases the acidity. An additional resonance structure would make this most clear.

The phrasing of the question really gives it away(!).  Look at the lower right resonance structure in 5.a. (above) - the one with the para negative charge.  A nitro group on the para carbon provides additional delocalization of the anion by withdrawing the electron pair, as shown below.  More resonance structures = greater stability; greater stability of a conjugate base = greater acidity of the proton.


 
 

6. (8 pts) As you know, primary amines react with aldehydes/ketones to form imine (C=N) products, instead of the alcohols typically seen for other nucleophilic addition reactions. However, this is not observed with secondary/tertiary amines or with alcohols (or any other nucleophile, for that matter). Why is this? Please use methylamine and acetone as your example reactants. You must include at least one mechanism in your answer, or it will not be graded.

Attack of neutral nucleophiles always involves loss of a proton from the Nu: (to maintain the correct number of bonds to it); the best example is an oxygen Nu: like H2O.  This would also be true with a 2o Nu: like Me2NH, and is illustrated in the first step below (the one forming the alcohol).  However, primary nucleophiles like methylamine have two protons to lose.  So, the second one can be abstracted, with loss of the adjacent OH (via an E2 reaction) to give the C=N double bond.


 

7. (4 pts) Rank the following molecules in terms of decreasing basicity, i.e. most basic to least basic, by placing each structure's letter in the appropriate space.

most basic =
   C
   E
   D
   A
   F
   B
 =least basic
Aromatic amines are weaker than aliphatic, so C/E are first.  B is last since it's quaternary and thus there is no electron pair.  C has two e--releasing Me's, while E has only one.  D has an e--releasing OMe, A has nothing, and F has an e--withdrawing NO2.

 
Bonus (5 pts, no partial credit) The usual protocol for nitration of aromatic rings involves slowly adding the substrate to a mixture of nitric and sulfuric acids (as we did in lab). For aniline, this proceeds in high yield to provide, primarily, the expected p-nitroaniline. In another procedure, the compound is dissolved in sulfuric acid, and the nitric acid is then added slowly. This also works fine for aniline, except that the main product is m-nitroaniline. This phenomenon is outlined in the box. Explain this mechanistically!

In the first case, the amino group is o,p directing for the usual reason - it has a free electron pair on the nitrogen which it shares with the ring, especially in the ortho and para positions.  In the second case, initial dissolution of the amine causes protonation of the amino group to give the PhNH3+ salt.  There is now no lone pair to donate; in fact, we now have a fully positive nitrogen atom on the ring, which we have learned is a strong meta director.